Problem: A certain circle can be represented by the following equation. $x^2+y^2+6y-72=0$ What is the center of this circle ? $($
The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+6y-72&=0\\\\ x^2+y^2+6y&=72\\\\ x^2+(y^2+6y)&=72 \text{(grouping terms)}\\\\ x^2+(y^2+6y{+9})&=72{+9}\end{aligned}$ Notice that we must add ${9}$ on the right side of the equation, since we added it to the left side of the equation. [How did we get 9?] Writing the equation in standard form $\begin{aligned}x^2+(y^2+6y{+9})&=72{+9}\\\\ (x-0)^2+(y+3)^2&=81\\\\ (x-0)^2+(y-(-3))^2&=9^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(0,-3)$ and has a radius of $9$ units. Summary The circle is centered at $(0,-3)$. The circle has a radius of $9$ units.